Determining the unitary D uniquely from the constraint $D^TMD=M$ with diagonal $M$

Question

I've made an edit to this question and title.

Consider a unitary matrix $D$ which satisfy the relation $$D^TMD=M$$ where $M$ is a diagonal matrix with non-zero diagonal elements $m_1,m_2,m_3$.

I'm trying to prove that D is diagonal and its elements are $D_{ij}=\pm \delta_{ij}$ provided $m_1\neq m_2\neq m_3$.

Assume $$D=\begin{pmatrix}a & b & c\\ e & f & g\\p & q & r\end{pmatrix}$$ and $$M=\begin{pmatrix}m_1 & 0 & 0\\ 0 & m_2 & 0\\0 & 0 & m_3\end{pmatrix}.$$

Since $D$ is unitary $D^\dagger D=DD^\dagger=I$. Therefore, $$MD=D^*M$$ which gives $$a=a^*,b=b^*\frac{m_2}{m_1}, c=c^*\frac{m_3}{m_1}$$ and so on.

I think, I have exhausted all the conditions on D. But I'm not being able to show that the off-diagonal elements are zero and diagonal elements are $\pm 1$.

Answer 1

It shouldn't particularly matter what the eigenvalues/diagonal entries of $M$ are since $D$ is constrained to be unitary. What this means in particular is that since $D^TD = D^2 = I$, we have that $D$ is its own inverse and therefore has full rank. The unitary and diagonal conditions force all the diagonal entries of $D$ to be $\pm1$. If in general we assume a diagonal entry is of the form $a + bi$ we must have that

$$ (a + bi)^2 \;\; =\;\; (a^2 - b^2) + 2ab i \;\; =\;\; 1 $$

meaning that either $a$ or $b$ are zero but not both. If $b=0$ we simply have $a = \pm 1$. If $a =0$ we then have that $b = \pm i$ and the diagonal entry is $bi = \pm i^2 = \mp 1$.

Actually under further inspection this is simpler (assuming complex entries). Simply having the unitary condition means $D^*D = DD^* = I$, which means that a diagonal entry must satisfy $\overline{d_{ii}} = d_{ii}$ which means the diagonal entries must be real, and $d_{ii}^2 = 1$.

Answer 2

Let my first answer assuming the problem is entirely real-valued.

From $D^T M D = M$ and $D D^T = I$ (due to unitarity) we can conclude $M D = D M$ (multiply $D$ from the left). In element-wise notation this reads as $$m_j d_{i,j} = d_{i,j} m_i$$ for $i,j=1,2,\ldots,n$. Note that we can write this as $$d_{i,j}(m_j - m_i) = 0.$$ If $m_j \neq m_i$ for all $j\neq i$ this implies $d_{i,j} = 0 \; \forall \; i \neq j$. Therefore, $D$ is diagonal $ D = {\rm diag}\{d_{11}, \ldots, d_{nn}\}$. From the unitarity of $D$ it follows that $d_{ii}^2 = 1$. Collecting the results we have $D_{ij} = \pm \delta_{ij}$.

edit: In the complex-valued case, it depends on your notion of $D^T$. If it means conjugate transpose then from $D^T M D = M$ we again get $M D = D M$ which leads to $d_{i,j} = 0 \, \forall i \neq j$. For $d_{i,i}$ from the unitarity constraint we get $|d_{i,i}|^2 = 1$ which means $d_{i,i} = {\rm e}^{\jmath \varphi_i}$ where $\varphi_i$ is arbitrary. This means your claim is not really true (partially, the diagonality is true but the $\pm \delta_{i,j}$ fails).

If it is just a transpose then the diagonality fails as well. Here is a counterexample:

$$D = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & \jmath & 0 \\ \jmath & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{bmatrix}, \quad M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix}.$$ As you can easily verify, $DD^\dagger=I$ (i.e., $D$ is unitary) and $D^T M D = M$.